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Would the first inequality be \[5c+32b \le 805\]?
Okay...I'm not sure what to put for the second one...\(c+b\le 80\)?
why did you write 805 Instead of 80 ?
there are 805 square meters in the parking lot.
Wait. Yoo mean 85
I made a typo...I meant 805, my bad
Assuming your inequalities are correct, what do you think would be the next step?
Don't I need a third inequality? For prices?
oh wait...no...I use the prices to find the min/max! So f(x,y)=2c+6b?
Will there only be one intersection point then?
But, would max be at (65,15) so 15 buses?
There's a range of points Within a region that are possible, but there is only one correct pair of coordinates that will maximize the profit
Show we how you found that result.
I have no idea...I just chose the intersection point...I'm sorry, I'm having a hard time understanding these word problems...with only two inequalities to graph...
There's almost always four inequalities to graph and you usually use x and y for variables
When I dealt with the three inequalities I plugged in the intersection points into the f(x,y) function...
Because you also consider .\[c \ge 0\] and \[b \ge 0\]
but If you use x and y as variables, then you can use a platform like desmos to plot your inequalities
That's what I'm doing :)
So, the intersection points I have now are: (80, 0) (65, 15) (0, 25.16) (0,0) But we can tell already that (0,0) will not be the maximum.
It is (65,15).
f(80,0)=160 f(0, 25.16)=150.96 f(0,0)=0 f(65,15)=220
Would you mind helping me with a couple more?
Sure, but you seem to already know what you are doing for the Most part.
I think for one I just need it checked...not sure about the last one..
Eh I'll just post the one I need help on :)